`lim_(x -> -oo) (x^3 - 4)/(x^2 + 1)` Find the limit.

Textbook Question

Chapter 3, 3.5 - Problem 26 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

Divide both numerator and denominator by `x^2` :

`(x^3-4)/(x^2+1) = (x-4/x^2)/(1+1/x^2).`

When `x -> -oo,` `1/x^2 -> 0.` Therefore the limit is equal to

`(-oo - 0)/(1+0) = -oo.`

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scisser | (Level 3) Honors

Posted on

Plug in `-oo` everywhere you have x
`lim_(x->-oo)(-oo^3-4)/(-oo^2+1)=(-oo)/-oo `

Since you have , you can use L^Hopital's Rule and differentiate the numerator and denominator independently.

`lim_(x->-oo)(3x^2)/(2x)=-oo/-oo `

Use LH's Rule again

`lim_(x->-oo)6x/2=-oo `

Therefore,

the `lim_(x->-oo)(x^3-4)/(x^2+1)=-oo `

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